Question bank — Velocity and acceleration in SHM — v = ω√(A² − x²)

False — at $x=\pm A$ the root $\sqrt{A^2-x^2}=0$, so $v=0$; the particle is momentarily frozen while it reverses. Distance travelled has nothing to do with instantaneous speed.

False — $a=-{\omega }^{2}x$, and at $x=0$ that gives $a=0$. The centre is where speed peaks but acceleration vanishes; they are anti-phase.

False — $v_{\max }=\omega A$ at the centre, $|a{|}_{\max }={\omega }^{2}A$ at the edges. They peak at opposite locations, which is exactly the $90°$/$180°$ phase story.

False — free fall has $a=g$ (fixed), but here $a=-{\omega }^{2}x$ changes with position: zero at centre, largest at the edges. Only $a\propto -x$ defines SHM, per Simple Harmonic Motion — definition a = −ω²x.

False — the bare root gives only the magnitude (speed). The physical velocity carries a $\pm$ because each point is passed once outbound and once inbound with opposite sign.

True — $v_{\max }=\omega A$ scales with $A$, but $\omega$ (hence $T=2\pi /\omega$) is set by $k$ and $m$, not by how far you pull it. See Angular frequency ω and time period T.

False — $v=\omega \sqrt{A^2-(A/2{)}^2}=\omega A\sqrt{3}/2\approx 0.87v_{\max }$. Speed stays high through the middle and drops steeply only near the edge; half-speed happens at $x=\sqrt{3}A/2\approx 0.87A$.

True — the minus sign in $a=-{\omega }^{2}x$ always drives the particle back toward equilibrium, so every displacement is reversed; remove the minus and the motion runs off exponentially instead of cycling.

The particle never reaches $x=2A$: SHM lives entirely in the range $-A\le x\le A$ because $x=A\sin (\omega t+\varphi )$ and $\sin$ can never exceed $1$. So $x=2A$ is outside the physical domain; the root turns imaginary precisely as a flag that you've asked about a displacement the motion can't produce, not because a real speed is complex.

The arithmetic is right but the conclusion isn't: when $x<0$, $a=-{\omega }^{2}x>0$ points in the $+x$ direction — which, per our sign convention, is back toward centre. "Restoring" means "opposite sign to $x$," and the minus sign guarantees that on both sides.

We did not differentiate w.r.t. $x$; we eliminated time using ${\sin }^2+{\cos }^2=1$ from Pythagorean identity sin² + cos² = 1. Differentiating $v$ gives acceleration, not the $v(x)$ relation.

They're identical. First, the stiffness–frequency link is $k=m{\omega }^2$ (the defining SHM force $F=-kx=ma=-m{\omega }^{2}x$), so $k/m={\omega }^2$. Second, the total energy equals its all-potential value at the turning point $x=A$ where $v=0$: $E=\frac{1}{2}k{A}^2$, hence $2E/m=(k/m)A^2={\omega }^2{A}^2$. Substitute both: $\sqrt{{\omega }^2{A}^2-{\omega }^2{x}^2}=\omega \sqrt{A^2-x^2}$. Same statement, energy language (Energy in SHM — kinetic and potential).

The last number is wrong: $a$ leads $v$ by $90°$, and $90°+90°=180°$, so $a$ leads $x$ by $180°$ (anti-phase), not $45°$. Phases add, not average — see Phase and phase difference.

The projected (shadow) speed is fastest as the point crosses the horizontal diameter (shadow at centre, $x=0$), and zero at the extreme left/right (shadow at $\pm A$). See Reference circle (projection of uniform circular motion).

Because $v=\omega \sqrt{A^2-x^2}$: the square root flattens near $x=0$ and plunges near $x=\pm A$, where the two nearly-equal squares cancel. Geometrically the "room left" shrinks like a semicircle's height, not a straight line.

It answers "how fast here?" directly, without first solving for the time the particle reaches that point — you skip an inverse-sine step and any phase bookkeeping entirely.

$a$ depends only on where you are ($x$) and how stiff the oscillator is (${\omega }^2$); the amplitude enters only through the largest $x$ the motion reaches, giving $|a{|}_{\max }={\omega }^{2}A$ at $x=\pm A$.

Because $v=\frac{dx}{dt}$: differentiating $\sin$ gives $\cos$, which peaks a quarter-cycle earlier. The rate of change always leads the quantity itself, so $v$ leads $x$, never lags.

Acceleration is fixed entirely by position (one value of $x$ → one $a$), so no ambiguity. Velocity is not: the same $x$ is visited with two opposite directions, hence the sign choice.

Take "outbound" as moving away from centre. On the $+x$ side, moving away means increasing $x$, so $v>0$ (the $+$ root); returning means $v<0$ (the $-$ root). On the $-x$ side it flips: outbound is decreasing $x$, so $v<0$, and inbound is $v>0$. The magnitude is the same both ways — only the direction (sign) changes.

$a=0$ means zero net restoring force at that instant, but the particle is moving fastest and carries maximum momentum straight through it — no force needed to keep going, and a restoring force reappears the moment it moves off centre.

$v=\omega A$ (maximum speed) and $a=0$ (no restoring acceleration). Fastest but momentarily un-accelerated — it coasts through the centre.

$v=0$ (frozen at the turning point) and $|a|={\omega }^{2}A$ (maximum restoring acceleration) pointing back toward centre. Speed and acceleration swap roles versus the centre.

Then $x=0$ for all time, $v=\omega \sqrt{0}=0$, and $a=0$: no oscillation at all. A zero-amplitude "oscillator" just sits at equilibrium.

$v_{\max }=\omega A\to 0$ and $|a{|}_{\max }={\omega }^{2}A\to 0$: the motion becomes infinitely slow and the period $T=2\pi /\omega \to \infty$. In the limit nothing moves in any finite time.

Yes — it correctly gives $v=0$ there. The turning point is the one place where speed and the "room left under the root" both vanish, which is precisely why the particle reverses.